The sample space is a unit cube in A-B-C space. Probabilities are the totals of the volumes of the cube taken up.

The plane A+B+C=2 cuts off a pyramidal corner of the cube, with vertex at (1,1,1), in which the total exceeds 2, and since it never exceeds 3, the first digit is 2. (Ignore probability zero points and planes representing the total being exactly 3 or 2, or .9, etc.) The base of the pyramid is an equilateral triangle with sides of length sqrt(2), as being diagonals of three of the cube's faces. The edges leading to the vertex of the pyramid are of length 1. The height of this pyramid is 1/3 the major diagonal of the cube, as the whole diagonal consists of equal segments going from (0,0,0) to the plane A+B+C=1, and then to the plane A+B+C=2 and finally to the plane A+B+C=3. Since the major diagonal of the cube is sqrt(3), the height of the pyramid is sqrt(3)/3. The area of the base is sqrt(3)/2. The volume of the pyramid is therefore (3/6)/3 = 1/6, this then being the probability that A+B+C > 2, and is therefore one of the volumes belonging to leading digit 2. (Remember that the probability of being exactly 2 or 3 is zero.)

The probability that A+B+C < 1 is equal to the volume of a congruent pyramid, this time with the vertex at (0,0,0). As its volume is also 1/6, this leaves the probability that A+B+C lies between 1 and 2 as 1 - 2/6 = 2/3, which is thus part of the overall probability that the first significant digit is 1.

This leaves the pyramid of A+B+C<1 to be divided up by leading digit.

Again, the pyramid has an equilateral triangle of side length sqrt(2) as the base, and has a height of sqrt(3)/3, and volume of 1/6.

The slice of this pyramid representing .9 (that is, the first of the slices that represent the digit 9) is a frustum of height sqrt(3)/30 (i.e., 1/10 the pyramid's height). The frustums for .8, .7, ... .1, are also of height sqrt(3)/30, but of decreasing base area.

The volume of that full pyramid is, as mentioned, 1/6. Without the slice representing .9, it's a similar pyramid, but with all linear dimensions equalling only 9/10 that of the original, so its volume is (9/10)^3 / 6, so the frustum representing .9 has volume (1 - (9/10)^3) / 6 = 271/6000.

This leaves a volume of 1/6000 (i.e., 1/1000 of the original pyramid) for the next iteration of the nine leading digits.

Each successive round then has 1/1000 the volume in the preceding round, applicable to each of the digits. So to account for successive 1/10 values (such as .09, .009, .0009, etc. for the digit 9), the values above need to be multiplied by 1000/999. This simply changes the denominators from 6000 to 5994. And we need to add 1/3 to the value for 2 to account for totals above 2 and 1/6 to the value for 1 to account for the totals between 1 and 2.